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Math Help - Inequality

  1. #1
    Junior Member
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    Inequality

    Hi !

    Either x,y,z> 0 , x + y + z = 1 . Prove That :

     (1+ \frac{1}{x} ) (1+\frac{1}{y} ) (1+\frac{1}{z} ) \geq 64
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  2. #2
    Junior Member
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    Dec 2009
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    Hi !

    So, This is My Solution :

    (1+\frac{1}{x})(1+\frac{1}{y})(1+\frac{1}{z}) \geq 64
    \Longleftrightarrow (x+1)(y+1)(z+1) \geq 64xyz
    \Longleftrightarrow xy+yz+zx+2 \geq 63xyz
    by AM-GM :
    xyz \leq \frac{(x+y+z)^3}{27}
    \Longleftrightarrow 54xyz \leq 2 (1)
    And We Have : (x+y+z)( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} ) \geq 9
    \Longleftrightarrow xy+yz+zx \geq 9xyz (2)
    by add (1) and (2) We Have The result sought !
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  3. #3
    MHF Contributor

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    Very good.
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